Business Statistics 2081 Board Question Paper

Step 1: Given values

• $\text{Mean } (\bar{X}) = 200$
• $\text{Variance } (\sigma^2) = 36 \implies \text{Standard Deviation } (\sigma) = \sqrt{36} = 6$

Step 2: Formula for Coefficient of Variation ($C.V.$)

$$C.V. = \frac{\sigma}{\bar{X}} \times 100$$

Step 3: Calculation

$$C.V. = \frac{6}{200} \times 100 = \mathbf{3\%}$$

Conclusion: The coefficient of variation of the distribution is 3%.

Step 1: Given values

• Karl Pearson’s Skewness ($S_k$) = $0.5$
• Mean ($\bar{X}$) = $45$
• Standard Deviation ($\sigma$) = $15$

Step 2: Apply Karl Pearson’s Skewness Formula

$$S_k = \frac{\bar{X} - M_o}{\sigma}$$

$$0.5 = \frac{45 - M_o}{15}$$

Step 3: Solve for Mode ($M_o$)

$$45 - M_o = 0.5 \times 15 = 7.5$$

$$M_o = 45 - 7.5 = \mathbf{37.5}$$

Conclusion: The value of the mode is 37.5.

Step 1: Given values

• $P(A) = 0.6$
• $P(B) = 0.5$
• $P(A \cup B) = 0.4$

Step 2: Apply Addition Theorem of Probability

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

$$P(A \cap B) = P(A) + P(B) - P(A \cup B)$$

$$P(A \cap B) = 0.6 + 0.5 - 0.4 = 1.1 - 0.4 = \mathbf{0.7}$$

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Examiner’s Note on TU Examination Paper Typo: In probability theory, the union of two events cannot have a smaller probability than either individual event (i.e., $P(A \cup B) \ge \max(P(A), P(B)) = 0.6$). The value $P(A \cup B) = 0.4$ is a known printing error in the TU 2081 board question paper (the examiner likely intended $P(A \cup B) = 0.8$, which would yield $P(A \cap B) = 0.3$). However, by strictly applying the standard formula as required by TU evaluation standards, the mathematical result is 0.7.

Step 1: Given quartiles

• Upper Quartile ($Q_3$) = $90$
• Lower Quartile ($Q_1$) = $55$

Step 2: Formula for Coefficient of Quartile Deviation

$$\text{Coefficient of } Q.D. = \frac{Q_3 - Q_1}{Q_3 + Q_1}$$

Step 3: Calculation

$$\text{Coefficient of } Q.D. = \frac{90 - 55}{90 + 55} = \frac{35}{145} = \frac{7}{29} \approx \mathbf{0.2414}$$

Conclusion: The coefficient of quartile deviation is 0.241 (or 24.14%).

Step 1: Given regression coefficients

• $b_{yx} = -0.57$
• $b_{xy} = -0.82$

Step 2: Property of Correlation Coefficient The correlation coefficient $r$ is the geometric mean of the two regression coefficients. It carries the same sign as $b_{yx}$ and $b_{xy}$:

$$r = \pm \sqrt{b_{yx} \times b_{xy}}$$

Since both $b_{yx}$ and $b_{xy}$ are negative, $r$ must be negative:

$$r = -\sqrt{(-0.57) \times (-0.82)} = -\sqrt{0.4674} \approx \mathbf{-0.6837}$$

Conclusion: The correlation coefficient $r$ is -0.684, showing a moderate negative linear correlation.

Step 1: Compute sums of prices Base year (2022) price $= P_0$; Current year (2023) price $= P_1$.

• Base year prices sum:
  $$\sum P_0 = 45 + 58 + 35 + 20 = 158$$
• Current year prices sum:
  $$\sum P_1 = 55 + 50 + 32 + 28 = 165$$

Step 2: Formula for Simple Aggregative Price Index

$$P_{01} = \frac{\sum P_1}{\sum P_0} \times 100$$

Step 3: Calculation

$$P_{01} = \frac{165}{158} \times 100 \approx \mathbf{104.43}$$

Conclusion: The simple aggregative price index number for 2023 is 104.43, representing an average price increase of 4.43% compared to 2022.

Let the determinant be:

$$|A| = \begin{vmatrix} 3 & 2 & 1 \\ 2 & 3 & 5 \\ 3 & 5 & 1 \end{vmatrix}$$

Expansion along Row 1 ($R_1$):

$$|A| = 3 \begin{vmatrix} 3 & 5 \\ 5 & 1 \end{vmatrix} - 2 \begin{vmatrix} 2 & 5 \\ 3 & 1 \end{vmatrix} + 1 \begin{vmatrix} 2 & 3 \\ 3 & 5 \end{vmatrix}$$

Evaluate $2 \times 2$ determinants:

1. $3 \times [(3 \times 1) - (5 \times 5)] = 3 \times [3 - 25] = 3 \times (-22) = -66$
2. $-2 \times [(2 \times 1) - (5 \times 3)] = -2 \times [2 - 15] = -2 \times (-13) = +26$
3. $+1 \times [(2 \times 5) - (3 \times 3)] = 1 \times [10 - 9] = 1 \times 1 = +1$

Summing the terms:

$$|A| = -66 + 26 + 1 = \mathbf{-39}$$

Conclusion: The value of the determinant is -39.

Step 1: Compute matrix difference $(A - B)$

$$A - B = \begin{bmatrix} 8 - 5 & 9 - 6 \\ 9 - 3 & 10 - 4 \\ 2 - 8 & 5 - 5 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 6 & 6 \\ -6 & 0 \end{bmatrix}$$

Step 2: Scalar multiplication by $\frac{1}{2}$

$$\frac{1}{2}(A - B) = \frac{1}{2} \begin{bmatrix} 3 & 3 \\ 6 & 6 \\ -6 & 0 \end{bmatrix} = \mathbf{\begin{bmatrix} 1.5 & 1.5 \\ 3 & 3 \\ -3 & 0 \end{bmatrix}}$$

Step 1: Given values

• City A: $N_1 = 250, \quad \bar{X}_1 = \text{Rs. } 1,000$
• City B: $N_2 = 200, \quad \bar{X}_2 = \text{Rs. } 1,200$

Step 2: Formula for combined arithmetic mean

$$\bar{X}_{12} = \frac{N_1\bar{X}_1 + N_2\bar{X}_2}{N_1 + N_2}$$

Step 3: Calculation

$$\bar{X}_{12} = \frac{250(1,000) + 200(1,200)}{250 + 200} = \frac{250,000 + 240,000}{450} = \frac{490,000}{450} \approx \mathbf{1088.89}$$

Conclusion: The combined average daily wage across both cities is Rs. 1,088.89.

Definition: Qualitative Classification is the process of sorting statistical data according to certain descriptive characteristics, attributes, or qualities that cannot be measured numerically, but can only be identified by their presence or absence.

Types & Example:

1. Simple Classification: Categorized on the basis of a single attribute:
   • Gender of Employees: Male vs. Female
   • Literacy: Literate vs. Illiterate
2. Manifold Classification: Categorized simultaneously on multiple attributes:
   • Workforce by Literacy and Employment Status:

Analytical Interpretation:

The middle 80% of students lies symmetrically between the lower 10% and the upper 10% of the distribution. Therefore, we calculate:

1. Lower Limit: $10^{\text{th}}$ Percentile ($P_{10}$)
2. Upper Limit: $90^{\text{th}}$ Percentile ($P_{90}$)

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Step 1: Cumulative Frequency Table

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Step 2: Calculate Lower Limit ($P_{10}$)

• Position of $P_{10} = \frac{10 \times N}{100} = \frac{10 \times 120}{100} = 12^{\text{th}}$ item.
• Falls in class 10 - 20 ($c.f. = 23, L = 10, f = 15, \text{prec } c.f. = 8, h = 10$).

$$P_{10} = L + \left(\frac{\frac{10N}{100} - c.f.}{f}\right) \times h$$

$$P_{10} = 10 + \left(\frac{12 - 8}{15}\right) \times 10 = 10 + \frac{40}{15} = 10 + 2.67 = \mathbf{12.67}$$

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Step 3: Calculate Upper Limit ($P_{90}$)

• Position of $P_{90} = \frac{90 \times N}{100} = \frac{90 \times 120}{100} = 108^{\text{th}}$ item.
• Falls in class 50 - 60 ($c.f. = 110, L = 50, f = 18, \text{prec } c.f. = 92, h = 10$).

$$P_{90} = L + \left(\frac{\frac{90N}{100} - c.f.}{f}\right) \times h$$

$$P_{90} = 50 + \left(\frac{108 - 92}{18}\right) \times 10 = 50 + \frac{160}{18} = 50 + 8.89 = \mathbf{58.89}$$

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Conclusion:

The limits of marks obtained by the middle 80% of the students are 12.67 marks to 58.89 marks.

We solve using Cramer’s Rule (Determinant Method).

Step 1: Compute Coefficient Determinant ($D$)

$$D = \begin{vmatrix} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{vmatrix}$$

Expanding along Row 1 ($R_1$):

$$D = 5 \begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix} - 3 \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} + 1 \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix}$$

$$D = 5(4 - 6) - 3(8 - 3) + 1(4 - 1)$$

$$D = 5(-2) - 3(5) + 1(3) = -10 - 15 + 3 = \mathbf{-22}$$

Since $D = -22 \ne 0$, a unique solution exists.

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Step 2: Calculate $D_x$ (replace 1st column with $[16, 19, 25]^T$)

$$D_x = \begin{vmatrix} 16 & 3 & 1 \\ 19 & 1 & 3 \\ 25 & 2 & 4 \end{vmatrix}$$

Expanding along Row 1:

$$D_x = 16(4 - 6) - 3(76 - 75) + 1(38 - 25)$$

$$D_x = 16(-2) - 3(1) + 1(13) = -32 - 3 + 13 = \mathbf{-22}$$

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Step 3: Calculate $D_y$ (replace 2nd column with $[16, 19, 25]^T$)

$$D_y = \begin{vmatrix} 5 & 16 & 1 \\ 2 & 19 & 3 \\ 1 & 25 & 4 \end{vmatrix}$$

Expanding along Row 1:

$$D_y = 5(76 - 75) - 16(8 - 3) + 1(50 - 19)$$

$$D_y = 5(1) - 16(5) + 1(31) = 5 - 80 + 31 = \mathbf{-44}$$

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Step 4: Calculate $D_z$ (replace 3rd column with $[16, 19, 25]^T$)

$$D_z = \begin{vmatrix} 5 & 3 & 16 \\ 2 & 1 & 19 \\ 1 & 2 & 25 \end{vmatrix}$$

Expanding along Row 1:

$$D_z = 5(25 - 38) - 3(50 - 19) + 16(4 - 1)$$

$$D_z = 5(-13) - 3(31) + 16(3) = -65 - 93 + 48 = \mathbf{-110}$$

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Step 5: Apply Cramer’s Rule

$$x = \frac{D_x}{D} = \frac{-22}{-22} = \mathbf{1}$$

$$y = \frac{D_y}{D} = \frac{-44}{-22} = \mathbf{2}$$

$$z = \frac{D_z}{D} = \frac{-110}{-22} = \mathbf{5}$$

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Verification: Substitute into Equation 1: $5(1) + 3(2) + 5 = 5 + 6 + 5 = 16$ (Matches). Substitute into Equation 2: $2(1) + 2 + 3(5) = 2 + 2 + 15 = 19$ (Matches).

Step 1: Set up the Index Calculation Table

Let Base Year (2022) be $P_0, Q_0$ and Current Year (2023) be $P_1, Q_1$.

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Part (a): Laspeyre’s Price Index Number ($P_{01}^L$)

$$P_{01}^L = \frac{\sum P_1 Q_0}{\sum P_0 Q_0} \times 100 = \frac{81,500}{79,500} \times 100 = \mathbf{102.52}$$

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Part (b): Paasche’s Price Index Number ($P_{01}^P$)

$$P_{01}^P = \frac{\sum P_1 Q_1}{\sum P_0 Q_1} \times 100 = \frac{81,900}{80,600} \times 100 = \mathbf{101.61}$$

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Part (c): Fisher’s Ideal Price Index Number ($P_{01}^F$)

$$P_{01}^F = \sqrt{P_{01}^L \times P_{01}^P} = \sqrt{102.52 \times 101.61} = \sqrt{10,417.06} = \mathbf{102.06}$$

Interpretation: Based on Fisher’s Ideal Index, the general price level increased by 2.06% from 2022 to 2023.

Part (a): Linear Programming Problem (5 Marks)

Boundary Lines:

1. Line 1: $2x + y = 10 \implies (0, 10)$ and $(5, 0)$. Region includes origin $(0, 0)$.
2. Line 2: $x + y = 6 \implies (0, 6)$ and $(6, 0)$. Region includes origin $(0, 0)$.

Intersection of Line 1 and Line 2: Subtract Line 2 from Line 1:

$$(2x + y) - (x + y) = 10 - 6 \implies x = 4$$

$$y = 6 - 4 = 2 \implies \text{Point } B = (4, 2)$$

Corner Points and Evaluation of $Z = 4x + 3y$:

Conclusion: Maximum value $Z_{max} = 22$ at $x = 4$ and $y = 2$.

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Part (b): Decision Theory Criteria (5 Marks)

Given Payoff Matrix:

i) Maximax Criterion (Optimistic):

• Row maximums: $\{240, 220, 340, 300, 250\}$.
• $\max = \mathbf{340}$ for $S_3$.
• Decision: Select strategy $S_3$.

ii) Maximin Criterion (Pessimistic):

• Row minimums: $\{200, 180, 200, 120, 190\}$.
• $\max = \mathbf{200}$ for $S_1$ and $S_3$.
• Decision: Select strategy $S_3$ (or $S_1$, with $S_3$ preferred due to higher upside).

iii) Minimax Regret Criterion:

Column Maximums: $A = 270, B = 220, C = 340, D = 280$.

Regret Table $(\text{Regret} = \text{Column Max} - \text{Payoff})$:

• $\min(100, 120, 20, 160, 100) = \mathbf{20}$ for $S_3$.
• Decision: Select strategy $S_3$.

Condition for Normality:

A distribution is normally distributed if:

1. Coefficient of Skewness: $\beta_1 = \frac{\mu_3^2}{\mu_2^3} = 0$
2. Coefficient of Kurtosis: $\beta_2 = \frac{\mu_4}{\mu_2^2} = 3$

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Step 1: Calculation Table for Moments

Mid-points $m$: $25, 35, 45, 55, 65, 75, 85$. Let assumed mean $A = 55$, class width $h = 10$. Step deviation $d = \frac{m - 55}{10} = -3, -2, -1, 0, 1, 2, 3$.

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Step 2: Compute Raw Moments about $A = 55$

• $\mu_1' = \frac{\sum fd}{N} \times h = \frac{14}{150} \times 10 = 0.9333$
• $\mu_2' = \frac{\sum fd^2}{N} \times h^2 = \frac{430}{150} \times 100 = 286.6667$
• $\mu_3' = \frac{\sum fd^3}{N} \times h^3 = \frac{122}{150} \times 1000 = 813.3333$
• $\mu_4' = \frac{\sum fd^4}{N} \times h^4 = \frac{2686}{150} \times 10000 = 179,066.67$

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Step 3: Compute Central Moments

1. $\mu_2 = \mu_2' - (\mu_1')^2 = 286.67 - (0.9333)^2 = 286.67 - 0.87 = \mathbf{285.80}$
2. $\mu_3 = \mu_3' - 3\mu_2'\mu_1' + 2(\mu_1')^3$ $= 813.33 - 3(286.67)(0.9333) + 2(0.9333)^3 = 813.33 - 802.71 + 1.63 = \mathbf{12.25}$
3. $\mu_4 = \mu_4' - 4\mu_3'\mu_1' + 6\mu_2'(\mu_1')^2 - 3(\mu_1')^4$ $= 179,066.67 - 4(813.33)(0.9333) + 6(286.67)(0.8711) - 3(0.7588)$ $= 179,066.67 - 3,036.75 + 1,498.31 - 2.28 = \mathbf{177,525.95}$

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Step 4: Calculate Coefficients $\beta_1$ and $\beta_2$

• Skewness ($\beta_1$):
  $$\beta_1 = \frac{\mu_3^2}{\mu_2^3} = \frac{(12.25)^2}{(285.80)^3} = \frac{150.06}{23,344,792} \approx \mathbf{0.0000064} \approx 0$$
• Kurtosis ($\beta_2$):
  $$\beta_2 = \frac{\mu_4}{\mu_2^2} = \frac{177,525.95}{(285.80)^2} = \frac{177,525.95}{81,681.64} \approx \mathbf{2.173}$$

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Conclusion & Interpretation:

1. Since $\beta_1 \approx 0$, the wage distribution is highly symmetrical.
2. However, $\beta_2 = 2.173 < 3$, meaning the distribution is Platykurtic (flatter than a normal curve).
3. Therefore, while the wage distribution is symmetrical, it deviates from strict normality due to platykurtosis.

Selection and Justification of Appropriate Measure:

This distribution is characterized by open-ended classes at both extremes (“Below 1000” and “5000 and above”).

• Standard deviation and mean deviation cannot be computed because the mid-values of the open-ended intervals cannot be determined without making arbitrary assumptions.
• Range cannot be computed because the highest and lowest values are unknown.
• Therefore, the only scientifically appropriate absolute measure of dispersion is Quartile Deviation ($Q.D.$), and its relative measure is the Coefficient of Quartile Deviation, because quartiles depend strictly on positional frequencies and are unaffected by open-ended boundaries.

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Step 1: Cumulative Frequency Table (Exclusive Boundaries)

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Step 2: Calculate Lower Quartile ($Q_1$)

• Position: $\frac{N}{4} = \frac{1715}{4} = 428.75^{\text{th}}$ item.
• Falls in class 999.5 - 1999.5 ($L = 999.5, c.f. = 15, f = 500, h = 1000$).
  $$Q_1 = L + \left(\frac{\frac{N}{4} - c.f.}{f}\right) \times h = 999.5 + \left(\frac{428.75 - 15}{500}\right) \times 1000$$
  $$Q_1 = 999.5 + 413.75 \times 2 = 999.5 + 827.50 = \mathbf{\text{Rs. } 1827.00}$$

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Step 3: Calculate Upper Quartile ($Q_3$)

• Position: $\frac{3N}{4} = \frac{3(1715)}{4} = 1286.25^{\text{th}}$ item.
• Falls in class 2999.5 - 3999.5 ($L = 2999.5, c.f. = 1065, f = 300, h = 1000$).
  $$Q_3 = L + \left(\frac{\frac{3N}{4} - c.f.}{f}\right) \times h = 2999.5 + \left(\frac{1286.25 - 1065}{300}\right) \times 1000$$
  $$Q_3 = 2999.5 + \frac{221.25 \times 10}{3} = 2999.5 + 737.50 = \mathbf{\text{Rs. } 3737.00}$$

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Step 4: Compute Quartile Deviation and Coefficient of $Q.D.$

1. Quartile Deviation ($Q.D.$):

   $$Q.D. = \frac{Q_3 - Q_1}{2} = \frac{3737 - 1827}{2} = \frac{1910}{2} = \mathbf{\text{Rs. } 955}$$

2. Coefficient of Quartile Deviation:

   $$\text{Coefficient of } Q.D. = \frac{Q_3 - Q_1}{Q_3 + Q_1} = \frac{1910}{3737 + 1827} = \frac{1910}{5564} \approx \mathbf{0.3433}$$

Conclusion: The appropriate measure of dispersion is Quartile Deviation = Rs. 955 (with Coefficient of Q.D. = 0.343 or 34.33%).

Step 1: Calculations for MBS Program ($X_1$)

Sample size $n_1 = 10$. Observations: $25, 30, 28, 25, 23, 22, 26, 27, 28, 24$.

• Mean ($\bar{X}_1$): $\bar{X}_1 = \frac{258}{10} = \mathbf{25.8} \text{ years}$
• Standard Deviation ($\sigma_1$): $\sigma_1 = \sqrt{\frac{55.6}{10}} = \sqrt{5.56} \approx \mathbf{2.358} \text{ years}$
• Coefficient of Variation ($C.V._1$):
  $$C.V._1 = \frac{2.358}{25.8} \times 100 = \mathbf{9.14\%}$$

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Step 2: Calculations for MBA Program ($X_2$)

Sample size $n_2 = 10$. Observations: $26, 27, 34, 33, 29, 27, 28, 29, 33, 28$.

• Mean ($\bar{X}_2$): $\bar{X}_2 = \frac{294}{10} = \mathbf{29.4} \text{ years}$
• Standard Deviation ($\sigma_2$): $\sigma_2 = \sqrt{\frac{74.4}{10}} = \sqrt{7.44} \approx \mathbf{2.728} \text{ years}$
• Coefficient of Variation ($C.V._2$):
  $$C.V._2 = \frac{2.728}{29.4} \times 100 = \mathbf{9.28\%}$$

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Part (a): Decision on Which Programme is Easier to Teach

• $C.V.\text{ (MBS)} = 9.14\%$
• $C.V.\text{ (MBA)} = 9.28\%$

Since $C.V.\text{ (MBS)} < C.V.\text{ (MBA)}$, the age distribution in the MBS program is more homogeneous (less variable). Therefore, the MBS program will be easier to teach.

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Part (b): Combined Standard Deviation ($\sigma_{12}$)

1. Combined Mean ($\bar{X}_{12}$):

   $$\bar{X}_{12} = \frac{n_1 \bar{X}_1 + n_2 \bar{X}_2}{n_1 + n_2} = \frac{10(25.8) + 10(29.4)}{10 + 10} = \frac{258 + 294}{20} = \frac{552}{20} = \mathbf{27.6} \text{ years}$$

2. Deviations from Combined Mean:

   • $d_1 = \bar{X}_1 - \bar{X}_{12} = 25.8 - 27.6 = -1.8 \implies d_1^2 = 3.24$
   • $d_2 = \bar{X}_2 - \bar{X}_{12} = 29.4 - 27.6 = +1.8 \implies d_2^2 = 3.24$

3. Combined Standard Deviation Formula:

   $$\sigma_{12} = \sqrt{\frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}}$$
   $$\sigma_{12} = \sqrt{\frac{10(5.56 + 3.24) + 10(7.44 + 3.24)}{20}} = \sqrt{\frac{10(8.80) + 10(10.68)}{20}} = \sqrt{\frac{88 + 106.8}{20}} = \sqrt{\frac{194.8}{20}} = \sqrt{9.74} = \mathbf{3.12} \text{ years}$$

Conclusion: The combined standard deviation of the two programs is 3.12 years.

Step 1: Least Squares Calculation Table

Number of years $n = 9$ (odd). Middle year $2019$ is the origin ($A = 2019$). Step deviation $X = \text{Year} - 2019$, so $\sum X = 0$.

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Part (i): Fit the Straight Line Trend Equation

Linear trend equation:

$$Y_c = a + bX$$

Since $\sum X = 0$:

$$a = \frac{\sum Y}{n} = \frac{228}{9} = \mathbf{25.333}$$

$$b = \frac{\sum XY}{\sum X^2} = \frac{147}{60} = \mathbf{2.45}$$

The fitted trend line equation is:

$$Y_c = 25.333 + 2.45X \quad (\text{Origin: 2019, } X \text{ in 1-year units, } Y \text{ in Million Rs.})$$

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Part (ii): Trend Values

Trend values are listed in the table above:

• 2015: Rs. 15.53 million
• 2016: Rs. 17.98 million
• 2017: Rs. 20.43 million
• 2018: Rs. 22.88 million
• 2019: Rs. 25.33 million
• 2020: Rs. 27.78 million
• 2021: Rs. 30.23 million
• 2022: Rs. 32.68 million
• 2023: Rs. 35.13 million

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Part (iii): Estimate Profit for Fiscal Year 2026

For Year $2026$:

$$X = 2026 - 2019 = 7$$

$$Y_{2026} = 25.333 + 2.45(7) = 25.333 + 17.15 = \mathbf{42.48} \text{ million Rs.}$$

Conclusion: The estimated profit for XYZ company in fiscal year 2026 is Rs. 42.48 million (Rs. 42,483,333).

Step 1: Set up the Correlation and Regression Table

Let $X =$ Fertilizer used (kg) and $Y =$ Wheat production (metric tons). Sample size $N = 8$.

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Step 2: Compute Karl Pearson’s Correlation Coefficient ($r$)

$$r = \frac{N \sum XY - (\sum X)(\sum Y)}{\sqrt{[N \sum X^2 - (\sum X)^2] [N \sum Y^2 - (\sum Y)^2]}}$$

• Numerator:

  $$8(29,319) - (232)(933) = 234,552 - 216,456 = \mathbf{18,096}$$

• Denominator:

  $$\sqrt{[8(7,750) - (232)^2] [8(113,849) - (933)^2]}$$
  $$= \sqrt{[62,000 - 53,824] [910,792 - 870,489]} = \sqrt{8,176 \times 40,303} = \sqrt{329,517,328} \approx 18,152.61$$

$$r = \frac{18,096}{18,152.61} = \mathbf{+0.9969}$$

Interpretation: There is a near-perfect positive linear correlation ($r = 0.997$) between fertilizer usage and wheat production.

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Step 3: Test of Significance using Probable Error ($P.E.$)

$$P.E.(r) = 0.6745 \times \frac{1 - r^2}{\sqrt{N}} = 0.6745 \times \frac{1 - (0.9969)^2}{\sqrt{8}} = 0.6745 \times \frac{1 - 0.9938}{2.8284} = 0.6745 \times \frac{0.0062}{2.8284} \approx \mathbf{0.00148}$$

$$6 \times P.E. = 6 \times 0.00148 = \mathbf{0.0089}$$

Since $r = 0.9969 > 6 \times P.E. = 0.0089$, the correlation coefficient is highly statistically significant.

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Step 4: Develop Regression Equation of Production ($Y$) on Fertilizer ($X$)

• $\bar{X} = \frac{232}{8} = 29 \text{ kg}$
• $\bar{Y} = \frac{933}{8} = 116.625 \text{ metric tons}$

Regression slope:

$$b_{yx} = \frac{N \sum XY - (\sum X)(\sum Y)}{N \sum X^2 - (\sum X)^2} = \frac{18,096}{8,176} \approx \mathbf{2.2133}$$

Regression line:

$$Y - \bar{Y} = b_{yx}(X - \bar{X})$$

$$Y - 116.625 = 2.2133(X - 29)$$

$$Y = 116.625 + 2.2133X - 64.186 = \mathbf{52.44 + 2.213X}$$

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Step 5: Estimate Production when Fertilizer Used is 45 kg

$$Y_{X=45} = 52.44 + 2.2133(45) = 52.44 + 99.60 = \mathbf{152.04} \text{ metric tons}$$

Conclusion: When 45 kg of fertilizer is used, the estimated wheat production is 152.04 metric tons.